Chapter I (Ca) vs. I (Zn)? (Sr) vs. I (Ba) vs. I (Ra)? 1.11 I2 of some period 4 elements?

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1 Chapter I(Ca) vs. I(Zn)? The first ionization energies of calcium and zinc are 6.11 and 9.39 ev, respectively (see Appendix 1). Both of these atoms have an electron configuration that ends with 4s 2 : Ca is [Ar]4s 2 and Zn is [Ar]3d 10 4s 2. An atom of zinc has 30 protons in its nucleus and an atom of calcium has 20, so clearly zinc has a higher nuclear charge than calcium. Remember though, it is effective nuclear charge (Z eff ) that directly affects the ionization energy of an atom. Since I(Zn) > I(Ca), it would seem that Z eff (Zn) > Z eff (Ca). How can you demonstrate that this is as it should be? The actual nuclear charge can always be readily determined by looking at the periodic table and noting the atomic number of an atom. The effective nuclear charge cannot be directly determined, i.e., it requires some interpretation on your part. Read Section 1.6, Penetration and shielding, again. Study the trend for the period 2 p-block elements in Table 1.3. The pattern that emerges is that not only Z but also Z eff rises from boron to neon. Each successive element has one additional proton in its nucleus and one additional electron to balance the charge. However, the additional electron never completely shields the other electrons in the atom. Therefore, Z eff rises from B - Ne. Similarly, Z eff rises through the d block from Sc - Zn, and that is why Z eff (Zn) > Z eff (Ca) I(Sr) vs. I(Ba) vs. I(Ra)? The first ionization energies of strontium, barium, and radium are 5.69, 5.21, and 5.28 ev, respectively (see Table 1.6). Normally, atomic radius increases and ionization energy decreases down a group in the periodic table. However, in this case I(Ba) < I(Ra). Study the periodic table, especially the alkaline earths. Notice that Ba is eighteen elements past Sr, but Ra is thirty-two elements past Ba. The difference between the two corresponds to the fourteen 4f elements between Ba and Lu. As explained above in the answer to Exercise 1.9, Z eff rises with each successive element because of incomplete shielding. Therefore, even though radium would be expected to have a larger radius than barium, it has a higher first ionization energy because it has such a large Z eff I 2 of some period 4 elements? The second ionization energies of the elements calcium - manganese increase from left to right in the periodic table with the exception that I 2 (Cr) > I 2 (Mn). The electron configurations of the elements are: Ca [Ar]4s 2 Sc [Ar]3d 1 4s 2 Ti [Ar13d 2 4s 2 V [Ar]3d 3 4s 2 Cr [Ar]3d 5 4s 1 Mn [Ar]3d 5 4s 2 Both the first and the second ionization processes remove electrons from the 4s orbital of these atoms, with the exception of Cr. In general, the 4s electrons are poorly shielded by the 3d electrons, so Z eff (4S) increases from left to right and I 2 also increases from left to right. While the I 1 process removes the sole 4s electron for Cr, the I 2 process must

2 remove a 3d electron. The higher value of I 2 for Cr relative to Mn is a consequence of the special stability of half-filled subshell configurations Ground-state electron configurations? (a) C? [He]2s 2 2p 2. (b) F? [He]2s 2 2p 5. (c) Ca? [Ar]4s 2. (d) Ga 3+? Since it is a cation there is no doubt that E(3d) < E(4s), therefore [Ar]3d 10. (e) Bi? Twenty-nine elements past Xe, which ends period 5 leaving the 5d and the 4f subshells empty, therefore [Xe]4f 14 5d 10 6s 2 6p 3. (f) Pb 2+? Only twenty-six electrons, past Xe, which ends period 5 leaving the 5d and the 4f subshells empty, therefore [Xe]4f 4 5d 10 6s More ground-state electron configurations? (a) Sc? [Ar]3d 14 s 2. (b) V 3+? Two electrons, past Ar, and since it is a cation there is no doubt that E(3d) <E(4s), therefore [Ar]3d 2. (c) Mn 2+? Five electrons, past Ar, therefore [Ar]3d 5. (d) Cr 2+? Four electrons, past Ar, therefore [Ar]3d 4. (e) Co 3+? Six electrons, past Ar, therefore [Ar]3d 6. (f) Cr 6+? Six elements past Ar, but with a +6 charge it has the same electron configuration as Ar, which is written as [Ar]. Sometimes inorganic chemists will write the electron configuration as [Ar]3d 0 to emphasize that there are no d electrons for this d-block metal ion in its highest oxidation state. (g) Cu? Eleven elements past Ar, but its electron configuration is not [Ar13d 9 4s 2. The special stability experienced by completely filled subshells causes the actual electron configuration of Cu to be [Ar]3d 10 4s 1. (h) Gd 3+? Seven electrons, past Xe, which ends period 5 leaving the 5d and the 4fsubshells empty, therefore [Xe]4f 7.

3 1.14 More ground-state electron configurations? (a) W? Twenty elements past Xe, fourteen of which are the 4f elements. If you assumed that the configuration would resemble that of chromium, you would write [Xe]4f 14 5d 5 6s 1. It turns out that the actual configuration is [Xe]4f 14 5d 4 6s 2. The configurations of the heavier d- and f-block elements show some exceptions to the trends for the lighter d-block elements. (b) Rh 3+? Six electrons, past Kr, therefore [Kr]4d 6. (c) Eu 3+? Six electrons, past Xe, which ends period 5 leaving the 5d and the 4fsubshells empty, therefore [Xe]4f 6. (d) Eu 2+? This will have one more electron than Eu 3+. Therefore, the ground state electron configuration of Eu 2+ is [Xe]4f 7. (e) V 5+? Five elements past Ar, but with a 5 + charge it has the same electron configuration as Ar, which is written as [Ar] or [Ar]3d 0. (f) Mo 4+? Two electrons, past Kr, therefore [Kr]4d I 1, A e, and C for period 3? The following values were taken from Tables 1.6, 1.7, and 1.8: Element Electron I (ev) A e C Configuration (ev) Na [Ne]3s Mg [Ne]3s Al [Ne]3s 2 3p Si [Ne]3s 2 3p P [Ne]3s 2 3p S [Ne]3s 2 3p Cl [Ne]3s 2 3p Ar [Ne]3s 2 3p In general, I, A e, and X all increase from left to right across period 3. All three quantities reflect how tightly an atom holds on to its electrons, or how tightly it holds on to additional electrons. The cause of the general increase across the period is the gradual increase in Z eff, which itself is caused by the incomplete shielding of electrons of a given value of n by electrons with the same n. The exceptions are explained as follows: I 1 (Mg) > I 1 (Al) and A e (Na) >A e (Al): both of these are due to the greater stability of 3s electrons relative to 3p electrons; A e (Mg) and A e (Ar) < 0: filled subshells impart a special stability to an atom or ion (in these two cases the additional electron must be added to a higher energy subshell (for Mg) or shell (for Ar)); I 1 (P) > I 1 (S) and A e (Si) > A e (P) the loss of an electron from S and the gain of an additional electron by Si both result in an ion with a half-filled p subshell, which, like filled subshells, imparts a special stability to an atom or ion.

4 1.16 Metallic radii of Nb and Ta? If you look at the elements just before these two in Table 1.4, you will see that this is a general trend. Normally, the period 6 elements would be expected to have larger metallic radii than their period 5 vertical neighbors; only Cs and Ba follow this trend: Cs is larger than Rb and Ba is larger than Sr. Lutetium, Lu, is significantly smaller than yttrium, Y, and Hf is just barely the same size as Zr. After Nb and Ta, the normal expectation is observed. There are no intervening elements between Sr and Y, but there are fourteen intervening elements, the lanthanides, between Ba and Lu. A shielding by the 4f electrons. For a more detailed discussion of this concept, study section 1.8(a), Atomic and ionic radii Electronegativities across period 2? Plots of electronegativity across period 2 and ionization energies across period 2 are superimposed on the figure below. The general trend is the same in both plots; both X and I increase from left to right across a period, and this is because the effective nuclear charge increases for the n = 2 orbitals across period 2. The two deviations in the upper plot result from different phenomena. For boron, the outermost electron occupies a 2p orbital, which has a higher energy than a beryllium atom s 2s orbital. The higher energy of the 2p electron offsets a boron atom s greater nuclear charge. For oxygen, two electrons are paired in one of the 2p orbitals, and the mutual repulsion they experience offsets an oxygen atom s greater nuclear charge relative to a nitrogen atom (see Example 1.5). Even though this exercise did not require the use of electron affinity values (Table 1.7), it is useful to think about the connections between the various atomic properties. Note that the electron affinities of beryllium and nitrogen are negative, and the explanation for these apparent anomalies is the same as the explanation given above for the departure of I 1 (B) and I 1 (O) from the general upward trend Frontier orbitals of Be? Recall from Section 1.8(c), Electron affinity, that the frontier orbitals are the highest occupied and the lowest unoccupied orbitals 3 of a chemical species (atom, molecule, or ion). Since the ground-state electron configuration of a beryllium atom is 1s 2 2s 2, the frontier orbitals are the 2s orbital (highest occupied) and the 2p orbitals (lowest unoccupied). Note that there can be more than two frontier orbitals if either the highest occupied and/or lowest unoccupied energy levels are degenerate.

5 1.19 Broad trends in I, atomic radius, and electronegativity? In general, as you move across a period from left to right, the ionization energy, I, 5 increases, the atomic radius decreases, and the electronegativity increases. All three trends are the result of the increase in effective nuclear charge, Z eff, from left to right.